Question

the mass number of iron nucleus is 56, the nuclear density is?

Answer 1

Nuclear density is the density of the nucleus of an atom, averaging about 2.3×1017 kg/m3. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars.

The nuclear density for a typical nucleus can be approximately calculated from the size of the nucleus, which itself can be approximated based on the number of protons and neutrons in it. The radius of a typical nucleus, in terms of number of nucleons, is {\displaystyle R=A^{1/3}R_{0}} where {\displaystyle A} is the mass number and {\displaystyle R_{0}} is 1.25 fm, with typical deviations of up to 0.2 fm from this value. The density of the nucleus is thus:

{\displaystyle n={\frac {A}{{4 \over 3}\pi R^{3}}}}

The density for any typical nucleus, in terms of mass number, is thus constant, not dependent on A or r, theoretically:

{\displaystyle n={\frac {A}{{4 \over 3}\pi (A^{1/3}R_{0})^{3}}}={\frac {3}{4\pi (1.25\ \mathrm {fm} )^{3}}}=0.122\ (\mathrm {fm} )^{-3}=1.22\cdot 10^{44}\ \mathrm {m} ^{-3}}

The experimentally determined value for n is 0.16 fm−3.

The mass density is the product of n by the nuclear mass. The calculated mass density, using a nucleon mass of 1.67×10−27 kg, is thus:

{\displaystyle (1.67\cdot 10^{-27}\ \mathrm {kg} )(1.22\cdot 10^{44}\ \mathrm {m} ^{-3})=2.04\cdot 10^{17}\ \mathrm {kg} \cdot \mathrm {m} ^{-3}}
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